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Chapter 9: Problem 59

A circular saw blade with radius \(0.120 \mathrm{~m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of \(2.00 \mathrm{rev} / \mathrm{s}^{2} .\) After the blade has turned through \(155 \mathrm{rev},\) a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of \(0.820 \mathrm{~m}\) to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Short Answer

Expert verified
The horizontal distance covered by the piece is obtained by substituting the known values into the range equation.

Step by step solution

01

Convert angular acceleration and displacement to linear

Use the relationships that link angular and linear quantities. For linear velocity \(v\), this is done by multiplying the angular velocity by the radius: \(v = \omega r\). The same rule applies for acceleration: \(a = \alpha r\). In this case, we convert the given angular acceleration (in rev/s\(^2\)) to rad/s\(^2\) by multiplying with \(2\pi\). This gives \(a = \alpha r = 2 \cdot 2\pi r\). Similarly convert the angular displacement of 155 revolutions to radians by multiplying by \(2\pi\). This gives a displacement, \(d = 155 \cdot 2\pi r\).
02

Find the final velocity

Use the equation of motion that relates displacement, initial velocity, acceleration and final velocity: \(v_f^2 = v_i^2 + 2ad\). Here initial velocity \(v_i\) is zero (starts from rest), acceleration \(a\) is the one found in step 1 and displacement \(d\) is also found in step 1. Solve for \(v_f\) to get the final (tangential) velocity of the blade and hence the initial speed of the piece.
03

Establish the time of flight

Use the kinematic equation for vertical motion: \(h = v_{iy}t + 0.5g t^2\), where \(v_{iy}\) is initial vertical velocity, 0 in our case since the piece departs horizontally; \(h\) is the height from which the piece fell, given as 0.820m, \(g\) is the acceleration due to gravity, and \(t\) is the time it takes for the piece to hit the ground. Solve for \(t\) to find the time of flight.
04

Calculate horizontal distance covered

The horizontal motion is uniform, so the horizontal distance or range \(R\) covered by the piece is given by: \( R = v_{ix} t\), where \(v_{ix}\) is the initial horizontal velocity, equal to the final tangential velocity of the blade found in Step 2 because the piece broke off tangentially, and \(t\) is the time of flight determined in Step 3. Solve for \(R\) to get the answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Motion
Angular motion describes how an object rotates around a fixed point or axis. It is characterized by quantities such as angular velocity and angular acceleration. These quantities define how quickly and how much the object speeds up during its rotation.
  • **Angular Velocity**: This is the rate at which an object rotates, usually measured in revolutions per second or radians per second. In the exercise, the angular velocity is initially zero because the blade starts from rest.
  • **Angular Acceleration**: This is the rate of change of angular velocity, described as revolutions per second squared (rev/s虏). Here, it鈥檚 given as 2.00 rev/s虏.
  • **Displacement**: Measures how much rotation has occurred, in revolutions or radians. For the exercise, displacement is 155 revolutions, which needs to be converted to radians for calculation.
Understanding these concepts allows us to relate angular quantities to linear motion through the radius of rotation, as seen when converting angular acceleration to linear acceleration.
Kinematics
Kinematics is the branch of mechanics that deals with motion without considering the forces that cause it. Here we explore how kinematics applies to both linear and angular motion.
  • **Equation of Motion**: Kinematics uses equations to relate velocity, acceleration, time, and displacement. The primary equation used here, \(v_f^2 = v_i^2 + 2ad\), allows us to find the final linear velocity from rest, which is crucial in solving for how far the piece travels horizontally.
  • **Time of Flight**: Since the piece of the blade drops freely under gravity, we use the kinematic equation for vertical motion \(h = v_{iy}t + 0.5g t^2\) to determine the time it takes to hit the ground. This is essential for calculating horizontal travel.
  • **Vertical Displacement**: The vertical distance the piece travels is a key factor, given as 0.820 meters, providing the necessary height variable for our kinematic equation.
By mastering these kinematic principles, we can efficiently navigate projectile motion problems like the one presented in this exercise.
Circular Motion
In circular motion, an object revolves around a central point. The path of the motion is a circle, and various physical laws and formulas apply, involving the radius as a crucial factor.
  • **Tangential Velocity**: Comprehending circular motion requires understanding tangential velocity (\( v = \omega r \)). This represents the linear velocity at any point along the edge of the circular path. For our problem, this determines the initial horizontal speed of the piece as it breaks off the blade.
  • **Centripetal Acceleration**: Although not directly addressed here, it often plays a role in circular motion, meeting the requirements for objects to maintain a circular path. It's calculated using \( a_c = v^2/r \).
  • **Relation to Linear Motion**: Circular motion concepts convert to linear motion through the radius of the circle. In this way, angular displacement, velocity, and acceleration convert to linear equivalents that are directly applicable in projectile motion once the object leaves its circular path.
Leading to projectile path analysis, these circular motion insights are valuable in comprehending how rotational dynamics transition to linear dynamics.

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Most popular questions from this chapter

Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. \(\mathrm{P} 9.90\) ). The cone has mass \(M\) and altitude \(h .\) The radius of its circular base is \(R\).

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

\(\begin{array}{llll}\mathrm{In} & \mathrm{a} & \text { charming } & 19 \text { th-century }\end{array}\) hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating disk \(2.50 \mathrm{~m}\) in diameter (Fig. E9.18). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it. (a) At how many rpm must the disk turn to raise the elevator at \(25.0 \mathrm{~cm} / \mathrm{s} ?\) (b) To start the elevator moving, it must be accelerated at \(\frac{1}{8} g .\) What must be the angular acceleration of the disk, in rad/s \(^{2} ?\) (c) Through what angle (in radians and degrees) has the disk turned when it has raised the elevator \(3.25 \mathrm{~m}\) between floors?

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is \(60.0 \mathrm{~cm}\) long and has mass \(0.400 \mathrm{~kg}\). (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a \(60.0^{\circ}\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the \(\mathrm{V}\) at its vertex?
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